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If CH isn't true, this means $$2^{\aleph_0}=\aleph_2\:.$$ Does this imply: $$2^{\aleph_1}=\aleph_3$$ or does the CH hold only for $\aleph_0$ and $\aleph_1$? Thanks

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Why should $2^{\aleph_0} = \aleph_2$ if CH fails? – t.b. Sep 5 '12 at 9:18
up vote 25 down vote accepted

No, CH just says that $2^{\aleph_0}=\aleph_1$, so its failure just says that $2^{\aleph_0}\ne\aleph_1$. We know that $2^{\aleph_0}\ge\aleph_1$, but it’s consistent that $2^{\aleph_0}$ be any cardinal $\aleph_\alpha$ such that $\alpha\ge 1$ and $\operatorname{cf}\alpha>\omega$. In particular, $2^{\aleph_0}$ could be $\aleph_n$ for any $n>1$.

If $2^{\aleph_0}=\aleph_2$, $2^{\aleph_1}$ would have to be at least $\aleph_2$, but it could be much bigger. For instance, it’s consistent that $2^{\aleph_0}=\aleph_2$ and $2^{\aleph_1}=\aleph_{17}$.

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The negation of the Continuum Hypothesis does not entail that there is only one infinity strictly between the natural numbers and the real numbers. In fact, given any any infinite cardinal $\kappa$ there is a cardinal $\lambda \geq \kappa$ such that $\text{ZFC} + 2^{\aleph_0} = \lambda$ is consistent (assuming the consistency of ZFC itself, of course).

Even if we assume that $2^{\aleph_0} = \aleph_2$, it could still be that $2^{\aleph_1} = 2^{\aleph_0} = \aleph_2$ -- this happens, for example, under the Proper Forcing Axiom, or more simply under the assumption of Martin's Axiom with $2^{\aleph_0} = \aleph_2$. (The Proper Forcing Axiom is a strengthening of Martin's Axiom -- well, formally a strengthening of $\text{MA} ({\aleph_1})$ -- which fixes the size of the continuum at $\aleph_2$.)

On the other hand, we can have $2^{\aleph_0} = \aleph_2$ and $2^{\aleph_1} > \aleph_3$. Very similar techniques for showing the consistency of $\neg \text{CH}$ can be applied to give the above.

The most general statement in this realm is Easton's Theorem which essentially says that except for certain basic restrictions, the function $\aleph_\alpha \mapsto 2^{\aleph_\kappa}$ can be consistently arbitrary.

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